stack.definition
stack.descriptionPart1
stack.descriptionPart2
stack.keyPointsTitle
- stack.keyPoint1
- stack.keyPoint2
- stack.keyPoint3
- stack.keyPoint4
stack.exampleTitle
push(30) โ stack.addedToTopstack.syntaxTitle
// Java
Stack<Integer> stack = new Stack<>();
stack.push(10);
int top = stack.pop();# Python
stack = []
stack.append(10)
top = stack.pop()// C++
#include <stack>
std::stack<int> s;
s.push(10);
int top = s.top(); s.pop();// JavaScript
const stack = [];
stack.push(10);
const top = stack.pop();Interactive Stack Visualization
enhanced.realWorldExamples.title
Browser history uses stacks to implement back and forward button functionality.
Why? LIFO order perfectly matches navigation history.
Text editors use stacks to track changes for undo/redo functionality.
Why? Stack enables efficient reversal of recent actions.
Time Complexity
Stack operations are highly efficient due to LIFO (Last In, First Out) access pattern. Push and Pop are O(1) since we always work with the top element.
| Operation | Time | Why? |
|---|---|---|
| Push | O(1) | Add element to top |
| Pop | O(1) | Remove element from top |
| Peek / Top | O(1) | View top element without removing |
| isEmpty | O(1) | Check if stack is empty |
| Size | O(1) | Get number of elements |
| Clear | O(n) | Remove all elements |
| Search | O(n) | Find element in stack |
enhanced.complexityAnalysis.title
enhanced.complexityAnalysis.mathematicalTitle
Push and pop operations are O(1) because they only modify the top pointer. No element shifting is required regardless of stack size.
๐ขBest Case
All operations (push, pop, peek) are O(1) in all cases.
๐กAverage Case
O(1) for push/pop operations. Search operations are O(n) as stack is not designed for random access.
๐ดWorst Case
O(1) for standard operations. Array-based stacks may need O(n) for resizing when capacity is exceeded.
๐พSpace Complexity
O(n) for n elements. Linked-list implementation has additional pointer overhead per element.
enhanced.codeWalkthrough.title
enhanced.codeWalkthrough.stepByStep
enhanced.practiceProblems.title
Valid Parentheses
easyGiven a string containing just '(', ')', '{', '}', '[' and ']', determine if the input string is valid. Brackets must close in the correct order.
"({[]})"trueContinue Learning
Test your knowledge with a quiz or explore implementation code in multiple languages.